could you please post an answer to this questionthanks A sam

could you please post an answer to this question.thanks

A sample of 75 students is selected and asked to provide their GPA. The mean GPA of the sample is calculated to be 2.26. The university indicated that the GPAs of all students is approximately normally distributed and over the past 10 years the standard deviation has been 0.41. What is the lower limit of the 98% confidence interval for the true mean of GPA of the entire university?

Solution

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=2.26
Standard deviation( sd )=0.41
Sample Size(n)=75
Confidence Interval = [ 2.26 ± Z a/2 ( 0.41/ Sqrt ( 75) ) ]
= [ 2.26 - 2.33 * (0.047) , 2.26 + 2.33 * (0.047) ]
= [ 2.15,2.37 ]